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AFubini's Theorem

Throughout this book we have concentrated on functions of a single real (oroccasionally complex) variable, other quantities such as the a in cos(ax) (Example 7.9) having the role of parameters. However in several places the variableshave a more equal role, notably when differentiating under the integral sign inthe proof of Lemma 7.52 or interchanging the order of integration to obtainParseval 's equation, Theorem 7.28. In order to justify these results we have tolook at functions of several real variables. We sketch as much of the theory aswe need, and refer the reader to [10] for a more systematic development.

To begin with notation, we reserve p, q for points in 1R2 with for instancep = (x , y) where x , y E lR. (Much of the appendix could equally be developed inIRn

, but two dimensions will be sufficient for our needs and will save on suffixes.)A function f is then (as in the introduction to Chapter 2) a rule which, foreach p in some subset of 1R2 assigns a real or complex number f (p) , the valueof f at p. We have to extend the concepts of continuity and differentiabilityfrom Chapters 2 and 3, and for this we need the notion of distance in 1R2 •

Definition A.I

Given points p = (x,y) , q = (u,v) E 1R2 , the distance d(p,q) = !p-ql isdefined as the positive square root

249

250

Definition A.2

Examples and Theorems in Analysis

A (real or complex valued) function 1 defined on E C ]R2 is said to be continuous on E if for each pEE and each e > 0 there is some 8 > 0 (depending onboth p and c) such that

11 (p) - 1 (q)1 < e for all q E E with Ip - ql < 8.

Another way of stating the definition is to say that with p = (x, y), then1(x + h, y + k) ---t 1 (x, y) as (h,k) ---t (0,0) from which it follows at once that1 must be continuous with respect to each variable separately, as consideredbelow. The dependence of 8 on p turns out to be a disadvantage, and weimmediately consider a stronger property.

Definition A.3

A (real or complex valued) function 1 defined on E C ]R2 is said to be uniformlycontinuous on E if for each e > 0 there is some 8 > 0 (depending on s only)such that

11(p) - l(q)1 < e for all p,q E E with Ip - ql < 8.

It is perhaps surprising that a function which is continuous on a set of theform [a , b] x [e, d] (a closed rectangle in ]R2) must be uniformly continuous there.We could have proved the corresponding result in Chapter 2, but chose on afirst acquaintance not to go beyond the results of Section 2.4. The analogues ofthose results are also true for functions on a closed rectangle, with essentiallythe same proofs.

Theorem A.4

Let E = [a ,b] x [e, d] and 1 be continuous on E. Then 1 is uniformly continuouson E.

Proof

Suppose that 1 is not uniformly continuous on E. Then for some c > 0 and every8 > 0 there are points p,q of E with Ip - ql < 8 and 11 (p) - 1 (q)1 2: c. (Thistests the reader's fluency with quantifiers!) Hence for each integer n 2: 1, wecan choose Pn, qn with IPn - qnl < lin and 11 (Pn) - 1 (qn)1 2: c. The sequence(Pn) is bounded, since it lies in [a, b] , and so by Corollary 1.29 has a convergentsubsequence, say (Pnj) whose limit, x say, is in [a, b] . Then for this subsequence,

A. Fubini's Theorem 251

(qnj) has again a convergent subsequence whose limit, y say, is in [c, d] . To avoidan excess of subscripts, now discard all terms not in the subsequence of (qn j)and relabel them as (Pr' qr) . This sequence of points has the properties that(Pr,qr) E E, Pr -+ x ,qr -+ y, IPr - qrl-+ 0 as r -+ 00, and If (Pr) - f (qr)1 ~ efor all r. But f is supposed continuous at u = (x, y) E E, so there is some6 > 0 such that If (u) - f (v)1 < e/2 for all vEE with Iv - u] < 6. Take rlarge enough so that IPr - xl , Iqr - yl < 6, when it follows that

If (Pr) - I (qr)1 < il (Pr) - I (u)1 + II (u) - I (qr)1

< e/2 + e/2 =e

and we have a contradiction. •

Despite its similarity to Definition 2.9, the new definition of continuity conceals some subtleties. For instance it is natural to consider one of the variablesx , y as fixed and allow the other to vary, thus giving the notion of continuityin each variable separately. However this is weaker than Definition A.2 as thenext example shows.

Example A.5

Let I (x, y) = xy/ (x2+y2) if (x, y) :I (0,0) , 1(0,0) = O. Then I is continuousin each variable separately at all points of]R2, but is not continuous at (0,0) .

To see this, note that f (x,y) = 0 whenever either x or y = 0, and so I iscontinuous in each variable separately at (0,0) . At other points it is continuousin each variable separately since xy and x2+ y2 are continuous and non-zerothere. However at points where x = y we have f (x,y) = 1/2, and since suchpoints may be arbitrarily close to (0,0) where f (0,0) = 0 it follows that f isnot continuous there.•

A similar subtlety arises when we come to differentiation. It is again naturalto consider fixing one of the variables, and differentiate with respect to theother, leading to the notion of partial derivatives as in the next definition.

Definition A.6

The limit

1. I (x + h,y) - I (x, y)1m h 'h_O

252 Examples and Theorems in Analysis

1. I (x ,y +k) - I (x, y)1m k 'k-+O

is the second partial derivative (the partial derivative with respect to the secondvariable - not to be confused with 021/ ay2 ) and denoted af/ay (x, y), etc .

if it exists, is called the first partial derivative of I at (x, y) (the partial derivative with respect to the first variable) and is denoted variously by

01ax (x, y), Ix (x, y), II (x, y) , or Dd (x , y) .

Similarly the limit

However the mere existence of partial derivatives at a point does not evenimply continuity in the sense of Definition A.2 since for instance Example A.5gives a function with partial derivatives at all points (in particular II (0,0) =h (0,0) = 0). For the definition of differentiability, we have, as in DefinitionA.2, to allow both variables to vary together.

Definition A.7

The function I is differentiable at (x, y) if there exist constants a,b and a function 8 (h, k) such that for all sufficiently small h, k, I (x +h, y +k) is definedand satisfies

I (x + h, y + k) = I (x, y) + ah + bk+ 8 (h, k) vlh2 + k2

where 8 (h, k) -> 0 as (h, k) -> (0,0) (equivalently 8 is continuous at (0,0) with8(0,0) = 0.)

It follows at once from the definition that a, b are the partial derivativesII, h at (x, y), and it also follows that if I is differentiable then it is continuous .Geometrically, Definition A.7 implies the existence of a tangent plane at (x, y)in the same way as Definition 3.1 implies the existence of a tangent line. Sincewe need only to prove the facts required in Chapter 7 we shall not develop thetheory further, but proceed at once to the main results .

Theorem A.8

Let I and h (the second partial derivative, a1/ay) be continuous on a rectangle[a, b] x (c, d) where a,b E lR and c, d may be finite or infinite. Then the functions¢ , 'l/J given by

¢(y) = l b

I(x,y)dx, 'l/J(y) = l b

h(x,y)dx

exist on (c,d) and ¢ is differentiable there with ¢' = 'l/J.

A. Fubini's Theorem

Proof

253

The integrals exist since 1,12 are continuous with respect to x . Given y E (c,d)we want to show that ¢' (y) = "IjJ (y) for all y E (c,d) . Choose a boundedinterval [cl ,dd c (c,d) with ci < Y < d1. Then I and 12 are continuous onE = [a ,b] x [Cl, d1] so by Theorem A.4 they are uniformly continuous there.Hence given e > 0 there is some 8 > 0 with 112 (p) - h (q)1 < E when p,q E Eand Ip - ql < 8. Then for k i 0 and y + k E let ,dd we have

¢ (y + kk - ¢ (y) - "IjJ (y) = lb{ I (x, y + kk - I (x , y) - h (x, y)} dx

b

= 1{h (x,y + k') - 12 (x,y)}dx

where 0 < k' [k < 1, using the Mean Value Theorem (for functions of onevariable). Hence if Ikl < 8, then

and the result follows.•

This version of the theorem is not the strongest known - it can with someeffort be extended to cover the case in which [a ,b] is replaced by an unboundedinterval - but it is easy to establish, and sufficient to prove Lemma 7.52 forinstance.

Finally we want Fubini's theorem which shows that under reasonable conditions , the value of a repeated integral is independent of the order in whichthe integrations are carried out.

Theorem A.9 (Fubini)

(i) Let I be continuous on E = [a, b] x [c, d]. Then h (x,y) = feY I (x, v) dv iscontinuous on E and

l (1' f (u, v) dU) dv ~ J.' h (u,d)du ~ J.'(l f (u,v) d+U

(A.l)

(ii) If I is continuous on I x J where I, J are intervals in R which may befinite or infinite , and if I is positive on E then the integrals in (A.1) areeither all infinite , or all finite and equal.


(iii) If I is continuous on I x J where I, J are intervals in R which may befinite or infinite, and if I is such that anyone of the intervals in (A.I)exists with III in place of I, then all are finite and equal.

Proof

We shall prove only (i); the other parts follow exactly as for double series inProposition 6.30 and Theorem 6.31.

The fact that h, as defined by h (x, y) = J: I (x ,v) dv, is continuous on E isimmediate from the uniform continuity of I - the details are left to the reader.Clearly 8hj8y = I (x, y) from the Fundamental Theorem of Calculus.

For y E Ie, d] , let

F(y) = lbh(X,y)dx=lb(lYI(X,V)dv)dx,

G(y) ~ 1'(1.' f(U,V)dU) dv

Again F,G are continuous functions on E, and we want to show that they arein fact equal. But F' (y) = J: 8hj8y (u, y) du = J: I (u, y) du by Theorem A.8

above, and G' (y) = J: I (u, y) du from the Fundamental Theorem of Calculus.Hence F (y) - G (y) must be constant, and since F (c) = G (e) the constantmust be zero. Hence in particular F (d) =G (d) which is the required result . •

We finish with an example to illustrate the failure of (iii) of the theoremwhen the integral of III is infinite.

Example A.ID

Let I (x , y) = sin (y - x) if x, y ~ 0, Iy - xl ::; 7T , = 0 otherwise. Then

The integral on the left is

l1r (l x

+1r

sin (y - x) dy) dx +100

(1~:1r sin (y - x) dY) dx

= -11r(Icos (y - x)l~+1r) dx -10C> (Icos (y - x)l:~:) dx

= - l 1r

(cos 7T - cosx) dx = 7T

A. Fubini's Theorem 255

and the integral on the right is the negative of this, on interchanging the rolesof x, y.•

BHints and Solutions for Exercises

It should go without saying that these hints and (partial) solutions are notintended to be read in parallel with the text; consult t hem only aft er makinga serious attempt to solve the question independently. For more on t his, andon the philosophy of 'learn ing t hrough problem solving,' see t he preface to [8].

Chapter 1

1. If x = a , y = -b with a > b > a then Ix+ yl = a - b < a while if b > a > athen Ix +yl = b - a < b. If z, y have the same sign, then this is also thesign of x + y.

2.ad - be

Sn+l - S n = (en +d)( e (n +1)+d)

so the sequence is increasing if and only if ad - be > O.

3. Any S n is an upper bound for S and so by Axiom 1.11, S has a least upperbound [ say, with [ :::; Sn for all n. Thus [ is a lower bound for (Sn). If[' > [ then some Sm < l' , since otherwise [' would be a lower bound for(sn) and thus an element of S, contrary to t he definition of l as t he leastupper bound of S. Thus [' is not a lower bound so l must be the greatest .

4. For some A , lao +aln +...+ak_lnk-11 :::; A nk- 1 and hence Sn ;::: aknkAnk- 1 which (i) is > a for n > Ajak so the sequence is bounded below,

257


and (ii) is > akn k/2 for n > 2A/ak and so the sequence is not boundedabove.

5. The limits are 3/4, 1/2, 1, 2/3 respectively.

6. Note that I/sni -ilil :::; ISn -ll . The converse is false; consider ((-It) .

7. If ISnl :::; m and e > 0, then there is N such that Itnl < e if n ~ N and so

ISntnl < me.

8. If Vo = 0, Vn+! = JV n +2 then 2 - Vn+l = 2 - JV n +2 = (2 - vn)/ (2 +~) < (2 - vn)/3, so 0 < 2 - Vn < (2 - vo)/3n as required.

9. Let r be the greatest lower bound of (r~/n) ; we show r~/n ~ r as n ~ 00 .

Given e > 0, there is some m with r;,(m < r+e. For any n, write n = km+tfor integers k , t with 0 :::; t :::; m-l. Then rn :::; rkmrt :::; r~rt :::; (r + e)km rt,and r~/n :::; (r +e)km/n ri/n. But km/n ~ 1 and ri/n both ~ 1 as n ~ 00,

so r~/n :::; r + 2e for large enough n, as required.

10. The sequence is increasing by induction, and bounded above by the positiveroot of the equation x 2 = ax + b.

11. Both an+!, bn+ l are strictly between an and b«, and an+! - bn+! =(~_ jb,02/2> O.

12. Suppose Sn ~ 1and tn = (SI + S2 + .. . + sn) In. Given e> 0, take m withISn - II < e for n ~ m. Then tn -l = (SI - 1+ S2 - 1+ .. . + Sn - 1) [n, sofor n ~ m ,

<

lSI -11 + ... + ISm - l -11 + .:..../S_m_-_l.:....1+_. ._.+--:..-Is~n_--..:.11n n

Const. n - m + 1 2--+ e « en n

for n large enough.

The example Sn = (-It for which tn ~ 1/2 shows that the converse fails .

13. Let n l / n = 1 + X n where X n > O. Then by the Binomial Theorem,

n = (1 + xnt > 1 + nXn + n (n -1) x~J2 > n (n - 1) x;/2, so x; <2/(n -1) ~ 0 as n ~ 00.

14. For k ~ 0, JSN+kl :::; ck ISNI ~ 0 as k ~ 00 .

15. Suppose for some x E [0,211"] , that in (x) = cosnx -> 1 as n -> 00. Thenin+l (x) + in-l (x) = 2cosnxsinx and taking a limit gives 21 = 21cosx ,so either 1= 0, or cos x = 1. But hn (x) = cos 2nx = 2 cos2 nx -1 and so 1must satisfy 1= 212 - 1 and in particular 1 :f: O. Hence for convergence we

B. Hints and Solutions for Exercises 259

must have cos x = 1, x = 2k1r for integer k when f n(x) = 1 for all n and soUn) converges. Hence cos nx has a limit if and only if cos x = 1, x = 2k1r.

If gn(X) = sin nx ~ mas n ~ 00, then gn+l(X)-gn-I(X) = 2 cos nxsinx ~oso either sin x = 0, or cos nx ~ 0 which we just found is impossible. Hencesin nx has a limit if and only if sin x = 0, x = k1r.

16. The condition implies that the sequence is bounded, and hence by Corollary1.29 it has a convergent subsequence with limit 1 say. Given e > 0 and Nwith 8n > s-« - e for n > m > N , let n ~ 00 while restricting 8n to thesubsequence. This gives 1 ~ 8 m - e, or 8 m ::; 1+£ for m > N. Since thesubsequence tends to 1we can choose 8 m in the subsequence with m > Nand 8 m > 1-£ and hence for n > m > N, 8n > 8 m-£ > 1- 2£. Combiningthese shows that the whole sequence tends to 1.

17. Immediate by induction.

18. If for some 0 > 0, tn ~ 0 > 0, then 8ntn ~ 8nO~ 00 . For examples to showthe failure of the converse, take for instance 8n = n2, tn = l/n, 1/n2, 1/n3 •

19. For Xn = 2, Xn+l = 2 - l/xn, the sequence is decreasing and > 1 byinduction, and the value follows from taking limits on both sides. Thesame argument works for any c > 2. What happens if c < 2?

20. (i) We have Xl = 1, X2 - Xl = 1/2, X4 - X2 > 1/2, ... ,X2k - X2k-1 > 1/2so X2k > 1 + k/2 follows by addition.

(ii) Yn+l-Yn = 1/(2n+1)+1/(2n+2)-1/(n+l) = 1/((2n+l)(2n+2)) > 0and Yn > n] (2n) = 1/2.

21. This follows by induction, since nl ~ 1, and nk+l > nk ~ k.

22. If 18n -11 < e for n ~ N then this also holds for those 8 n which are in thesubsequence.

23. (i) The result is obvious if L = 00, so suppose L finite, and let (8nk) be thesubsequence used in the proof of Theorem 1.27 which converges to L.Then the way in which (8nk) was constructed shows that if nk ::; j ::;

nk+l then 8j ::; 8nk when (8nk) is decreasing, 8j ::; 8nk+1 when (8nk) isincreasing. Hence any limit of terms 8j must be ::; lim 8nk = L.

(ii) That 8n > L - e for infinitely many n follows by putting n = nk. Also8n < L + £ if n = nk and k is large enough, and the inequalities notedin (i) show that then 8j < L + e if j ~ nk .

(iii) When (8nk) is increasing, sUPm~n s-« is eventually constant = L. If(8nk) is decreasing, then each sUPm~n 8m is equal to some peak point.


24. If the result is false. then from (iii) of the previous exercise, there is someN such that for all n :2: N

( al ::n+l)n< (1 + ~)n,equivalently

al an an+l-- < - - -- for n :2: N.n+1 n n+1

Adding these inequalities gives

( 1 1) aN an+lal N + 1 + ... + -; < N - n + 1

which is impossible since it gives an upper bound for 1/ (N + 1)+...+ l/nwhich we know tends to infinity.

(This striking result is from [7, Volume 1, pages 79-83], which gives anattractive discussion of the general approach to problem solving.)

25. The iterated limits are both 1, but the repeated limit does not exist sincefor instance the terms are zero when m = n .

26. For instance 5/7 occurs in the 7th , 14t h , 7kt h block of terms , while 20+5/7occurs in the 7kt h block when 7k :2: 21. Generally a fraction p/q in lowestterms occurs in the qkt h blocks when qk > pfq.

27. This exercise should bear a health warning! Verify by hand or machinethat the result is true up to at least 30, and plot the number of iterationsrequired to return to 1. Try another block of initial values around say 100.(Be warned that this has been done already up to many millions!) Try toexplain why there is a tendency for sets of adjacent initial values to haveequal numbers of iterations to return to zero. There is at the time of writingno convincing reason, heuristic or otherwise, why the result should be truein general.

Chapter 2

1. Suppose that f is increasing and has jumps at x , y with x < y and I =(f(x-),f(x+)) , J = (f(y-) ,f(y+)). Then if x < u < v < y thenf (u) ~ f (v) and so letting u -+ x+, v -+ v- , f (x+) ~ f (y-). Thus I , Jare disjoint. There can be only countably many such intervals since eachcontains a rational number by Theorem 2.39 and these numbers must bedistinct.


2. If x '/:- 1 then divide by x - 1 to get I (x) = x4 + x3 + x2 + X - 5 -+ -1 asx -+ 1. Hence k should be 1.

3. Let I be increasing, and suppose I maps (a,b) onto (c,d) . Let x E (a, b)and let e > 0 where (f (x) - e,I (x) + e) C (c, d) . Then there are u,v withu < x < v, I (u) = I (x) - e, I (v) = I (x) + e and so I maps (u, v) onto(f (x) - e, I (x) + e) which proves continuity.

4. One way is proved as Proposition 2.18. For the other suppose that I isnot cont inuous at c. Then there is some e > 0 such that for every 8 > 0there is some x with Ix - c] < 8 and II (x) - I (c)1 ~ e. For each n ~ 1,choose X n with IXn - cl < l/n and II (xn) - I (c)1 ~ e. Then X n -+ C butI (xn ) -.'-t I (c) which gives the converse.

5. Take e = II (c)1 /2 in the definition of continuity.

6. Let (an) , (bn) , c be as in the proof of the theorem, so that an -+ C-, b« -+

c+ . Since I is regulated at c there is some 8> 0 such that II (x) - I (c-)I <1 on (c-8,c) and I/ (x ) - / (c+)1 < 1 on (c,c+8) , so I is bounded on(c - 8, C + 8) and the proof is completed as before.

7. (an) is increasing, (bn) is decreasing, and bn - an -+ O.

8. For a > 0, IxQsin (l/x)1 ~ IxlQ-+ 0 as Ixl -+ O. For x> IT, sin (l/x) ~ l/x(Proposition 6.49(ii)) and so IxQsin (l/x)1 ~ Ix!Q-l -+ 0 as x -+ 00.

9. Ix2- a21 ~ [z - allx + al. However IJiXi -01 is not ~ M Ixl on (0,1) for

any M.

10. If x < u < Y < v, then I (x) ~ I (u) ~ I (y) ~ I (v) . Letting u -+ x , v -+ ythen I (x) s I (x+) s I (y) ~ I (y+) so I (x+) is monotone. If I iscontinuous on the right , then for any e > 0, there is some 8 > 0 witho~ I (y) - I (x) < e when 0 ~ y - x < 8. Then 0 ~ I (y+) - I (x+) ~ ewhen 0 ~ y - x < 8 follows as in the proof of the previous statement.

11. Let 9 (x) = I (x + 1/2)- I (x) . Then 9 is continuous and 9 (O)+g (1/2) = 0since 1(0) = 1(1). Thus either 9 (0) = 9 (1/2) = 0, when we are done, or9 (0) and 9 (1/2) have opposite signs and the result follows from the Intermediate Value Theorem. A similar argument, with 9 (x) = I (x + 1/3) I (x) gives the result with 1/3 (or generally l/n) in place of 1/2.

To show the result false for 9 (x) = I (x +c) - I (x) where 1/3 < c < 1/2,define I as follows. Take some d with 1/3 < d < c and let 1(0) = I (d) =1(1/2) = 1(1 - d) = 1(1) = O. Then define I on [1/2 ,1 - d] to be continuous and strictly positive on (1/2,1 - d) . Let m be an upper bound for Ion [1/2,1 - d] . Define I on [0, d] to be continuous and strictly positive on(0, d) with the additional requirement that I (x) > m on [1/2 - d,1 - 2d]


(not forgetting to verify that [1/2 - d,I- 2d] C (O ,d)) . For other values ofx, let I (x) = -I (1 - x). Show that this I does the trick, and then extendthe argument to cover other values of c =1= I/n.

12. Let's answer the second question first. Borrow from the proof of Theorem1.27 the idea of a peak point , so x E [a, b] is a peak point if I (x) 2': I (y)for all y 2': x . Thus G is the set of all x which are not peak points. Then ifwe imagine the sun rising from a distant point in the direction of positivex, G consists of the points which are in shadow and the result says that Ihas equal values at the end points of the intervals of shadow.

Suppose then that (c,d) C G, c, d </:. G, so c, d are peak points. ThenI (c) 2': I (d) since c is a peak point. If I (c) > I (d), I attains a maximumon [c,d] at some point k < d which must be a peak point since d is a peakpoint. Hence k </:. G which contradicts (c, d) C G. This leaves I (c) = I (d)as the only possibility.

13. (i) Clearly To (x) = 1, T1 (x) = x, T2 (x) = 2x2 - 1. Since cos (n + 1) () +cos(n -I)() = 2 cos n() cos () we have Tn +1 (x) + Tn - 1 (x) = 2xTn (x)which defines Tn inductively with the required properties. Notice thatin addition, as x = cos () decreases from 1 to -1 , () increases from °to1r and so cos n() changes monotonically from 1 to -1 and back n times,

finishing at (-1t when x = -1.

(ii) If p is a polynomial with the given properties, then the graph of p mustintersect the graph of Tn at least n times on the interval [-1,1] (onceon each branch from -1 to 1). Thus since p - Tn has degree n - 1 withn zeros, it must be identically zero . (The reader should check that theargument works when some of the zeros coincide.)

(iii) Ifq has degree n, leading coefficient 1, and Iq (x)1 :::; 21-n on [-1,1] then

q = 21- nTn from (ii); equivalently if q =1= 21-

nTn , then Iq (x) I > 21

-n

for some x E [-1 ,1] .

14.

I(x) = { :+ 1/2x -1/2

for x E Q,for 0 < x < 1/2, x</:. Q,for 1/2 < x < 1, x</:. Q.

15. (i) Rearranging as x = (5 - x)1/3 gives x = 1.51598.

(ii) Rearranging as x = tan- 1 x + 1r, gives x = 4.49341.

16. For any n 2': 0, IXn+l - dl = il (xn ) - I (d)1 :::; c IXn - dl ·

17. If I (x) -l l then I (x) - I (Y) -lOis immediate. Suppose conversely thatI (x) - I (Y) -l 0 as x,Y -l 00. Then if X n -l 00, I (xn ) is a Cauchy


sequence with a limit l , say, and the limit is independent of the choice of(xn) since if also Yn -4 00 we consider (Xl ,Yl, .. . ,Xn, Yn, . . .) . The rest ofthe argument is as in Exercise 4 above.

Chapter 3

1. The given functions are (a) continuous on (i) [0,00), (ii) JR, (iii) [0,00),(iv) [-1 ,1], and (b) differentiable on (i) (0,00), (ii) lR-7rZ, (iii) [0 ,00)(including on the right at 0), (iv) (-1,1) .

2. Suppose I' (e) < t < I ' (d) and consider 9 (x) = I (x) - tx . Then 9 isdifferentiable on [e, d] with g' (e) < 0 < g' (d) so 9 must attain a localminimum at some e E (e,d) at which g' (e) = O.

3. (i) Suppose I is differentiable at t, I' (t) = k and x < t < y. Then(J (t) - I (x)) / (t - x) -4 k as X-4 t - and (J (y) - I (t)) / (y - t) -4 kas y -4 t + . Also

I (y) - I (x) = t - x I (t) - I (x) + y - t I (y) - I (t)y-x y-x t-x y-x t-x

= c/(t)-I(x)+{3I(Y)-I(t)t-x t -x

where o, {3 > 0 and (X + {3 = 1. Hence (J (y) - I (x)) / (y - x) lies between (J (t) - I (x)) / (t - x) and U (y) - I (t)) / (y - t) and so must-4 k too .

(ii) Let X n = 1/ (n7r) so I (xn ) = (-It / (n27r2) • Show that

(-It (J (xn) - I (xn+1 )) / (xn - xn+d -42/7r.

4. (i) Given e > 0 there is some 8 > 0 such that If' (t) -ll < e for a < t <a + 8. Then for each x E (a,a + 8) there is some t E (a, x) such that(J (x) - I (a)) / (x - a) = I' (t) by the Mean Value Theorem. Hencefor each x E (a,a + 8) , !U (x) - I (a)) / (x - a) -ll < c as required.

(ii) If not, then for some p > m, I (b) - I (a) ~ p (b - a) (or < -p (b - a) inwhich case the argument is similar). Consider lines Y = px + e parallelto the chord on [a, b] and choose the least e for which the line intersectsthe curve. Let t be the greatest value of x which lies both on the lineand the graph of f. Then I~ (t) ~ p which is a contradiction.

(iii) The reader is left to modify the proof of (i) to cover this case.


5. (i) The tangent at (t,t2 ) is y - t2 = 2t (x - t) and this passes through(0, -1) if t = ±1 but never through (0,1) .

(ii) The normal at (t, t2 ) is 2t(y - t2 ) + (x - t) = 0 and this passes through(0, -1) if t = 0 only, through (0,1/2) if t = 0 (triple solution) andthrough (0,4) if t = 0, ±.j7/2.

(iii) The tangent at (t, et ) is y - et = et (x - t) which passes through theorigin if t = 1.

6. f (x) = a" (1 - x)13 is continuous on the right at 0 if a> 0 and on the leftat 1 if {3 > 0, and is always differentiable on (0,1) .

7. For f (x) = x2 , f' (c) = 2c = (f (b) - f (a» / (b - a) = (b+ a) /2, solimb_a (c - a) / (b - a) = 1/2. In fact the limit is 1/2 for any functionf and any point at which f" is continuous and non-zero, by consideringthe Taylor expansions of f and f' about a.

8. Suppose 1f'(x)1 :s; m on (a,b]. Then for a < x < b, If(x) -f(b)1 :s;mix - bl :s; m (b - a) and so f is bounded on (a,b] .

9. Suppose that f is defined and satisfies (3.20) on [c, b] - otherwise choosela', b'] c (a,b) and apply the solution on la', b'] .

(i) The statement in (a) is equivalent to (3.20). For (b) , suppose x < u < t,and f (u) is below the chord joining (t, f (t» to (y, f (y» . Then f (t) isabove the chord joining (u, f (u)) to (y, f (y)) which contradicts (3.20).

(ii) Continuity follows from (i) since for any x E (a,b) the graph off on (x , b) is between the the chord joining (a,f (a» to (x,f (z)and the chord joining (x,f(x)) to (b,f(b» . Also (3.20) shows that(f (t) - f (x» / (t - x) is decreasing as t ---+ x+ and bounded below bythe slope of the chord joining (a, f (a» to (x,f (x)) so the right-handderivative exists and is finite. A similar argument works on the left andshows that [': (x) :s; f~ (x) .

(iii) Choose u, v with x < u < v < y. Then (f(u) - f(x»/(u - x) :s;(f(y) - f(v))/(y-v) and so f~ (x) :s; l': (y) by taking limits. Hence if fis not differentiable at any points x, y then the intervals (J~ (x) ,f~ (x))and (J~ (y) ,f~ (y») are disjoint and non-empty, and so can be at mostcountable in number. (Compare the solution to Exercise 1, Chapter 2.)

(iv) is immediate from the Mean Value Theorem.

10.* (i) If f is not strictly increasing on (a,b) then there is (c,d) c (a,b) withf (c) 2: f (d). Then f cannot be constant on (c,d) since then f~ wouldbe zero on (c,d) which is an uncountable set . Hence either f (c) > f (d)


or, if I (c) = I (d) then there is some e E (c,d) with I (e) i: I (c) ; in anycase we have points p,q from among c,d,e with p < q and I (p) > I (q).

Now consider, for each y E (f (q),I (p)) (there are uncountably manysuch values of y), the greatest x with I (x) = I (y) . For such a point ,I (x) ;::: I (x') for x < x' < q and so I~ (x) ::; O. This gives uncountablymany points with I~ ::; 0 and a contradiction.

(ii) Apply (i) to I (x) + kx for arbitrary k > O.

11. Replace x by Y/3 to get a cubic with both I and I' having integer roots.

12. One needs to generate an infinite family of solutions I to the previousexercise, depending on a parameter which is chosen to make the roots ofJI also integers. But this has not been done.

13. The construction of (xn) , (Yn) shows that Yn < Yn+l < xn+! < Xn so thatYn -> l for some l and we know that Xn -> d where I (d) = O. Taking alimit in (3.19) gives l = d and so Yn < d < Xn for all n. We leave it to thereader to show that if f' ;::: m > 0 on (a,b) then Xn - Yn+l ::; I (xn) [ttiwhich gives convergence of (Yn) to d at the same rate as (xn) .

14. The definition of xn+! gives I(xn)/(xn - Xn+!) = f'(xn) and I(xn)/(xn d) = f'(u) for some u E (d,xn) by the Mean ValueTheorem . Hence (Xn+ld)/(xn - Xn+!) = f'(X n)/ f'(U) ::; f'(X n)/ f'(Yn).

15. Let In (x) = nx (1 - xt . Then In (x) -> 0 for each x E [0,1] but In attainsits maximum at Xn = 1/ (n + 1) and In (xn) = (n/ (n + l)t -> l/e i: O.

16. I (x) = x2 (x - 5) (x - 9) is zero at 0,5,9 and is > 0 on (-00,0), (0,5)and (9,00) , and < 0 on (5,9) . In particular it has a local minimum at O.f' (x) = 2x (x - 3) (2x - 15) and so since there must be a local maximumon (0,5) , it is at x = 3, and since there must be a local minimum on (5,9) ,it is at x = 15/2 and is a global minimum. There is no global maximum.

17. Ifa,b > 0 then I(x) = xa(l_x)b is continuous on [0,1]' positive anddifferentiable on (0,1) with 1(0) = 1(1) = O. The only stationary point isat x = a/ (a + b) which must be a maximum. If a,b < 0 then I -> 00 asx -> 0,1 with a unique minimum at a/ (a + b). If ab < 0 then f' has fixedsign on (0,1).

18. With I(x) = (sinx)/x, f'(x) = (xcosx-sinx)/x2 which is zero whentanx = x which happens exactly once in each interval (mr, (n + 1/2) 11') ,n ;::: 1. Similarly f" (x) = 0 when tanx = 2x/ (2 - x2

) which happensexactly once in each interval ((n - 1/2) 11',n1l') , n ;::: O.

19. The statement is the same except for replacing x -> a by x -> 00, and the


proof is got by replacing 0 < Ix - al < 8 by x > A (so x is 'near infinity ')and making the right verbal changes .

20.* One obvious example is when f is linear, f(x) = ax +b when (f(x + h) f(x))/h = a for all x, h. The limit does not exist for polynomials of degreegreater than 1, or for trigonometric or exponential functions , a" with a > 1.The limit does exist, and is equal to zero, when f(x) = lnx.

What is the relationship between the 'derivat ive at infinity' andlimx -+oo f'(x)?

21. If 9 = hf2 then by the product rule, Theorem 3.14,

9' Ii f~

9 = f1 + f2

which is the basis for an induction: if 9 = hf2 ... fnfn+1 then

9'

9=

Chapter 4

=

lTr

dx 100

dt= 2 -4o a +cos x - 0 a (1 +t2 ) + 1 - t 2

4 100

dt 2 a + 1= -- ~b2' where b = --1a-I 0 t + a-

4 [-1 ]00 27r(a - 1) b tan (t/b) 0 = vfa2=l '

1. (i) Put x = sinO to get Jo1y11-x2dx = JoTr/2cos20dO = JoTr/2(1

cos 20)/2 dO = 7r/4 (showing that the calculus and geometrical values for

the area of a quarter-circle are the same) . (ii) J12ln t [t dt = [ln2 t/2] r =

(In2 2) / 2. (iii) J;/4tanudu = J11/y'2dv/v(put v = cosu) = In(V2) =

(In 2) /2. (iv) Integrate by parts twice to get (7r2- 2b2 ) b- 3 sin (b7r) +

27rb- 2 cos (b7r). (v) Put tan (x/2) = t , dx = 2dt/ (1 + t2) , cosx =

(1 - t 2) / (1 + t2

) to get

t" dxJo a+cosx

(Of course in the last example we have been making free with results fromelementary calculus which will be justified in the next two chapters.)


2. A regulated function can only have jump discontinuities , i.e, points at whichthe right- and left-hand limits are unequal , either to each other or to thevalue of the function. But Exercise 2, Chapter 3 shows that if f is differentiable on [a, bJ then l' assumes on (a,b) every value between l' (a) andl' (b) (irrespective of whether l' is continuous) and so cannot have jumpdiscontinuities.

It is worth noting that the situation in quite different for one-sided derivatives: f (x) = Ixl has a right-derivative at every point, equal to -1 for x < 0and +1 for x 2: 0 - see Exercise 6 below.

3. These are immediate since by putting x = -t we have J~a f (x) dx =

Joaf (t) dt = J; f (x) dx if f is even, J~a f (x) dx = - Joaf (t) dt =- Joaf (x) dx if f is odd.

4.* For (i), f(t) = Xl +t(X2 -Xl), g(t) = Yl +t(Y2 - yt} for 0 ~ t ~ 1, andfor (ii), f(t) = r cos t, g(t) = rsint for 0 ~ t ~ 271" .

For the cycloid we have l' (t) = r (1 - cos t) , g' (t) = r sin t so the lengthis

r2121f V(I-cost)2+sin2tdt = r212'Tf ../2-2costdt

= 4r2121r sin (t/2) dt = 8r2.

5. (i) For n 2: 2,

{1f/2 {1f/2Jo sin" xdx = Jo sinn

-l x sinxdx

= [_ cosxsinn- l x]~/2 +l1f

/2

cosx (n - 1) sinn-

2 xcosxdx

r/2

= (n - 1) Jo sinn-

2 x (1 - sin2 x) dx = (n - 1) (In-2 - In)

as required. Then (4.11) is equivalent to ynIn = ../n - 2In- 2(n

1)/Jn(n - 2) and (ii) follows since n -12: In(n - 2).

Since In 2: In+1 we have ynIn 2: v'TiTIIn+1 In/ (n + 1) and so taking limits with n even gives a 2: b, and with n odd gives b 2: a. Multiplying the expressions in (4.11) gives 12nI2n+l = 71"/ (2 (2n + 1)), or.j21iI2n../2n + 1I2n+1 = (71"/2) J2n/ (2n + 1) and hence a = b = J7I"/2.

6. Let 9 (x) = J: r; We know from Theorem 4.13 that 9 is continuous on[a, b] and has a right-hand derivative g,+ which is equal to f+ on [a ,b),


using Exercise 4(iii), Chapter 3. Hence we have to show that if f, g arecontinuous on [a ,b] and g~ = f~ then f = g. Exercise 10(ii), Chapter 3 iswhat is required.

The reader is now challenged to use the full strength of that exercise toprove the following:

Theorem B.1

Let f be continuous on [a ,b] and suppose that f~ exists on [a, b] - D forsome countable set D. Define g (x) = f~ (x) on [a, b] - D, g (x) = 0 on D,and suppose that g is regulated. Then

l bg= f (b) - f (a) .

Chapter 5

1. If e < a' < a < b then I:' f + I:, f = I:' f + I: f + I: f = I: f + I: [,2. For 0 < a < 1, loa (1 - t2)-1/2 dt = J;in-

1

a dx = sin- 1 a, putting t = sinx.

Now let a - 1 - .

Similarly

l at (1 _ t2r1/2 dt = l sin

-1

a sinxdx = [_ cosx]~in-l a

= 1-~=a2f(1+~)-1

asa-1-.

We already have from Example 4.18 the integrals Ie-ax sin(bx)dx =e-ax(-a sin(bx)-b cos(bx))f(a2+b2), Ie-ax cos(bx)dx = e-ax(-a cos(bx)+bsin(bx))f(a2 + b2), so putting in limits O,e and letting e - 00 gives1000 e-axsin (bx) dx = b] (a2+ b2) , 1000 e-axcos (bx) dx = a] (a2+ b2) .

3. Let 4b - a2 = t 2 , t > O. Then (putting in the infinite limits directly) we


have

Joo dx

-00 x2+ax +b= 4Joo dx

-00 4x2+4ax +a2+ t2

= 21OO~ = ~ [tan-1 ¥.]oo-00 y2 + t 2 t t -00

211" 11"= t = -vr.b=-=a~2/774

The partial fractions (to be verified!) are

1 1 [x+v'2 x-v'2]x4 + 1 = 2v'2 x2+ v'2x +1 - x2 - v'2x +1

and so the integral is

4~ H::~~:::)L +HI:x2+~x+l100 dx ] 11" 2 11"

+ -00 x 2 - v'2x +1 = 4 VI - 1/2 = v'2.

The method for I~oo x2/ (x4 +1) dx is similar, and gives the same answer,

11"/v'2.

4. Since Iboo9 is finite , I: 9 ~ 0 as y, z ~ 00, so II: II ~ 0 also. Now use

Exercise 17, Chapter 2.

5. The function given by I (x) = n on tn, n +n-3 ) , n = 1,2,3, . .. , I (x) = 0otherwise, is unbounded and I~oo III = E~ n-2 which is finite.

The function given by I (x) = n2 on tn, n + n-3/2), = _n2 on [n +n-3/2,n+n-3 ) is integrable (improperly) since L(-lt /n is convergent,but I~oo III = L l/n diverges .

6. Put t4 = U (more generally tb = u) to reduce to a Beta-integral. For theintegral over (0,00) put tb = u, then u] (1 - u) = v.

7. I; sin (JX) dx = 2va cos (va) - sin (va) which is unbounded as a ~ 00,but if a = (2n + 1/2)211"2, the integral has the constant value -1.

8. We have to show that if loa I ~ l as a ~ 00 then subject to the givenconditions on g, 1000 I (x) 9 (tx) dx ~ last ~ 0 +. Suppose that e > 0 is

given, and that A > 0 is such that II:II < e for A :::; a < b. We first show

thatb1ts <eg(a) for A:::; a < b. (B.1)


Note that it is enough to show (B.1) when f ,9 are step functions (whyis this?) , and hence that it is enough to prove the analogue for sums: ifIL:~ arl ::; M for 1 :s: n :s: Nand (br) is decreasing and positive, then for1 :s: n :s: u,

n

L arbr :s: blM.I

But on rearranging we have with An = L:~ ar ,

n

L arbr = Albl + (A2 - AI) bz + ...+ (An - An-d b.,I

= Al (bl - b2) + ...+ An- l (bn- l - bn) + Anbn,n

L a.b; < M (bl - b2 + ...+ bn- l - bn + bn) = Mb,I

(B.2)

as required. Now returning to the proof of the main result , let A be as in(B.1), and let 8 > 0 be such that 9 (0) - 9 (tA) < c/A if 0 < t < 8. Then

lA

f (x) 9 (tx) dx -lA

f (x) dx s A (ciA)

which is enough, since we already know that I: f and I: f 9 are small.

For the examples we have the values 1000 e- tr sin (bx) dx = b] (t 2 + b2 ) ,

1000 e- t x cos (bx) dx = t] (t2 + b2 ) from Exercise 2 above. Hence letti ngt -+ 0+ gives the g-values 1000

cos (bx) dx = 0, 1000sin (bx) dx = lib for

real b f= O.

9. The idea of the solution is to show that if f does not tend to zero, thenusing the fact that f' -+ 0, there are intervals on which I f does not tendto zero, which contradicts the existence of 1000 f.

More explicitly, if f does not tend to zero, there is some 8 > 0 such thatIf (xn)1 ~ 8 for Xn -+ 00. Take c = 82/2; since f' -+ 0, there is some Nsuch that If' (x)1 :s: e if x ~ N and we assume Xn ~ N from now on. Thenon the interval [xn, xn + 1/8] we have If(xn)1 ~ 8, If' (x)l:S: c = 82/2andso If (x)1 ~ 8/2. Hence

t:X

n

f ~ (8/2) (1/ 8)

does not tend to zero, and so Jooo f cannot exist by Exercise 4 above.

B. Hints and Solutions for Exercises

Chapter 6

271

1. We have L:::'=o r" = (1 - r)-l for -1 < r < 1 from the Binomial Theorem,and so by differentiation (to be justified) L:::'=lnrn - 1 = (1 - r)-2 and soL::'=l nrn = r (1 - r)-2 . For k = 2,3, etc . differentiate again.

2. For instance if k = 3 we have

00 1 NIl N (1 1)~ n(n+3) = J~oo~ n(n+3) = 3J~oo~ ; - n+3

1 '. [( 1 1 1 ) (1 1 1= - lim 1+ - + - + . .. + - - - + - + -3 N -'00 2 3 N 4 5 6

+ . .. +_1)]N+3

1 [( 1 1) . (1 1 1)]=- 1+-+- - lim --+--+--3 2 3 N-.oo N + 1 N + 2 N + 3

1( 1 1)=3 1+2"+3

and the general pattern is clear.

3. Use the inequality

roo f(x)dx > tf(n) >100

f(x)dxJ2 3 3

and the given value of the integral.

4. If m = 2n is even, SS S2n+l = S2n + b2n + 1, so S- S2n S b2n + 1.

5. For each N = 1,2, ... , first take enough terms LPn for the partial sumto exceed 2N, then enough terms L qn for the partial sum to be less than2N - 1. The result can be further extended to make the rearrangementhave arbitrary limits superior and inferior (Exercise 23, Chapter 1).

6.* L:~ (-It diverges, but by grouping pairs we can get series (1 - 1) +(1 - 1) + ... which converges with sum zero, or 1- (1 - 1) - (1 - 1) - .. .with sum 1.

7.* The absolute convergence (to the correct sum) of the Cauchy productis a special case of Theorem 6.31. If an = (-1t Ivn then since y7;Jj S(a +b) 12, (-It L:;,:-J aran-r = L;':-1111 Jr (n - r) ~ L:;,:-J 2/n =2 (n - 1) In does not tend to zero.

8.** Some necessary conditions, and some (different!) sufficient conditions arein [4].


9.* The recurrence in (ii) follows from equating powers of t in

tetx = (et -1) t Bn ~x)tn. (B.3)n=O n.

This proves that the Bn are polynomials of degree n . Differentiating (B.3)with respect to x gives B~ (x) = nBn- 1 (x) and integrating this givesJ; s; (x) dx = 0 for n ~ 1.

10.* (i) is immediate from the previous question.

(ii) With x = 0 we have

te t - 1

t ~ B2n 2n= -2" +~ (2n)!t ,

~ B2n 2n= Lt (2n)!t ,

n=O

and putting t = 2iz gives the result .

11. Suppose2etx 00 t"

et + 1 = L En (x) n!n=O

and show, as in the previous question that En is a polynomial of degreenand ""n = 2nEn (1/2) is an integer (the nth Euler number) . Then x =1/2 , t = 2iz gives

secz

showing incidentally that all ""n with odd n are zero.

12.* (i) If f is positive and summable over A then for each integer n ~1 , theset En on which f (x) ~ l/n must be finite, otherwise LEn f (x) wouldbe infinite. Hence the set E on which f f= 0 is at most countable, andLEf (x) = LA f (x) is finite. Conversely if LA f (x) is finite then fcan only be > 0 on a countable set and f is summable over A. Therest follows by considering positive and negative parts (and real andimaginary parts if necessary), as in the proof of Proposition 6.24; inparticular if say g (x) , h (x) are the positive and negative parts andLg(x) is infinite, then choose E as the set on which g(x) > o.


(ii) The fact that the sums are either both finite or both infinite followssince~A/f (a)1 is an upper bound for ~:=l (~aEAn If (a)1) (over finitesubset of An) and conversely. The equality follows (inevitably?) byconsidering positive and negative parts.

(iii) For Theorem 6.27, A = Z+ and each An is the one-element set {n}; forTheorem 6.31, A = Z+ X Z+ and An is {n} x Z+ = {(n, 1), (n, 2), . . . ,(n, k), .. .} .

13. Put P= lim sup., .....oo lan!1/n , including the cases in which p = 0,00.

If p > 0, then given any e > 0, lanl > (p - s)" for infinitely many n.Then if Izi > 1/ (p - e), lanznl > 1 for infinitely many n and so ~ anzn

cannot converge. Hence all points at which the series converges must haveIzi ~ 1/ (p - e) , and so Izi ~ 1/Psince e is arbitrary. In particular if p = 00,the series converges only at z = 0.

If p < 00, then given any s > 0, there is some N such that lanl < (p + s)"for all n 2: N. Hence if Izi < 1/p we choose e such that Izi < 1/ (p + 2e) ,and so lanznl < ((p +e) / (p +2e)t for n 2: N, and the series is convergent. In particular if p = 0, the series converges for all z.

14. Let s« = It + h + ...+ [« . The construction shows that the maximum ofboth 81 and 82 is attained at x = 1/2 (and other places) when y = 1/2.Similarly the maximum of both 83 and 84 is attained at x = 1/2 -1/8 (andother places) when y = 1/2 + 1/8. Generally, the maximum of both 82n-1and S2n is attained at x = 1/2 - 1/8 - '" - 1/22n-1(and other places)when y = 1/2 + 1/8 + ... + 1/22n- 1 . Letting n --+ 00 (and anticipatingExercise 27 below), the maximum is at x = 1/3 when y = 2/3. For thevalue of f at other points, t say, write t = ~~ tn /2n with each t« E {0,1}(i.e. write t in binary) and for k 2: 1, let rk = 2:~=k tn / 2n - k. Show thatf (t) = 2:~ 21

-n It (rn ) .

15. For Izi ~ 1, [sinz] s 2:;;" Iz12n+1 / (2n + 1)! s jzl~;;" 1/ (2n + 1)! < 21zl .

Since the graph of sine is concave down on (0, 7r/2), the graph lies abovethe chord from (0,0) to (7r/2,1) which is what the result states.

If Izi s 1/2, [log(1 + z) - zl / Izl 2 = I~~ (_1)n-1 zn-2 /nl s ~~ 22-n/2=1.

16. For instance (abt = exp (xln (ab)) = exp(bxlna) = a(bx).

17. This is of course immediate from Example 6.77. Without going this far, theformulae (6.13) show that for instance sin z = °when exp (iz) = exp (-iz) ,or exp (2iz) = 1, which we know from Theorem 6.51(iii) is true only whenz is a multiple of tt .


18.* cos(x + iy) = cosx cos(iy) - sin x sin(iy) = cos x cosh y - i sin x sinh y. Fortan, rationalise the denominator.

19. Given z, write z = ea+i b with -7r < b ::; 7r so that log z = a + ib andexp log z = z. On the other hand. if z = x + iy then exp z = ex+iy and solog exp z = x+iyo where Yo = y+ 2k7ri and k is chosen to make Yo E (-7r , 7r].Thus log exp z = z if and only if y = Yo E (-7r ,7r].

20. If C = u + iv , f = 9 + ih then I cf = u I 9 - v I h + i (u I h + v I g) =(u + iv) (J9 + i I h) , etc .

Given f, choose c such that [c] = 1 and I cf > 0 (the result is trivial if

If = 0). Then II fl = II cfl = I cf = IlR(cf) s I Icfl = Ilfl ·21. For instance

00

L nr n-

1 cos nB =n=1

=

d ( I-rcosB )dr 1 - 2r cos B+ r2

cos () - 2r + r2 cos B2 .

(1 - 2rcosB + r2 )

The same results can be obtained by differentiation with respect to B (asshould be checked).

Letting r ~ 1- gives the g-sum -II (4sin2 (B/2)) for the (divergent)series L:=1 n cos nB when 0 < () < 27r.

22. Suppose Cl + C2 + '" + Cn < 1, (otherwise the result is trivial) and useinduction. For 1 ::; r < n ,

(1 - cI)(I - C2)'" (1 - cr )(1 - Cr+l)

2: (1 - (Cl + C2 + + cr))(1 - Cr+l)

= 1 - (Cl + C2 + + Cr + Cr+l) + (Cl + C2 + ... + Cr)Cr+l

> 1 - (Cl + C2 + +c; + cr+l) '

23. This follows by expanding all products and replacing each term by itsabsolute value.

24.


25.* (i) The binomial series for (1 - y2r 1/

2is uniformly convergent on every

interval (-1 + J, 1 - J) and so may be integrated termwise to get theseries for arcsin on (-1, 1) . At y = ±1, the series are convergent by theAlternating Series Test and we can take a limit as y -+ ±1 (anticipatingthe result of Exercise 28 below).

(ii) The series for arctan follows by integrating (1 + y2) - 1 as in (i), or moreinterestingly, by putting () = 1r/2 in (6.16) .

26. Since cos x ~ 1, I; cos t dt ~ I; dt, i.e. sin x ~ x. Then I; sin t dt s J; t dt,1 - cos x ~ x 2 /2. The general result follows by induct ion.

27. Let m be the maximum of f on [a, b] and let f (c) = m. Given e > 0 thereis N such that If (x) - fn (x)1 < e for all x E [a, b] and n ~ N.

Then fn (c) > f (c) - c for n ~ N so mn ~ m - e for n ~ N. Also for allx E [a, b], .f (x ) ~ f(c) < f(c) + c and so for n ~ N , [« (x ) ~ f(c ) + esince If (x) - t« (x)1< c. Hence mn ~ m +e for n ~ N as required.

28. The solution to Exercise 8, Chapter 5 concerning summability of integralsreduces to the case of series which is what we have here.

Since 2::~ r" = (l-r)-l and 2::'; nrn - 1 = (l-r)-2 for Irl ~ 1, then lettingr -+ 1- gives 1/2, 1/4 for the g-values of 2::~ (-It and 2::'; (-It- 1 n.Compare also Exercise 21 above .

Chapter 7

1. (i) The function is even , so we can use (7.4) to find (an) .

if n ~ 1.

ifn = 0

217< 21ax- f(x)cosnxdx = - (1- -) cos nx dx

1ro 1ro a

=~[x_~2]:=;

2 ]a 2 la= - [(1- =:) sinnx + - sinnxdx

n1r a 0 n1ra 0

= -i- (1 - cos na) = -:- sin2 (n2a)n 1ra n 1ra

The Fourier series is absolutely convergent for all x by the M-test , andso its sum is f as required.

for all n ~ O.


(ii) Here again, 1 is even and

2a (1l"na)= cos -1l" (1 - n 2a2 ) 2

(The reader should think about what happens to the term in whichn = k when a = 11k for some integer k.) Convergence follows as in (i).

2. The easiest way seems to be to use the fact that the series is convergentby Example 6.19 and apply the result of Exercise 28, Chapter 6.

3.* (i) To get started,

11

I/(t)Bdt)dt = 11f'(t)(t -~)dt

= [I (t) (t -~)J: -11

1 (t) dt

= 1 (0) ; 1 (1) -111 (t) dt.

Then the question guides you through the remaining steps .

(ii) From (i),

1 (j) +1 (j + 1)2 J

j +1 B1 (t) dt + L ~ [1(2k-l) (j + 1)

. (2k)1J 2::;2k<n '

( l)n 13+1-r:» (j)] + --=r . I(n) (t) Bn (t) dt,n. J

and the result follows by adding from j = m to n - 1.

4. For 0 < t < 21l" we have from Example 7.17,

1l" - t ~ sinnt-2- = L.J-

n-

n=1

which gives the case k = 1, putting t = 21l"x. Then integrate repeatedly,using f sin t dt = sin (t - 1l"12) .


For any k;::: 2,

277

kl 00 kl' ~ k ' ( k k )< 2--k L...J n: ~ 2--

k1 +2- +3- + . ..

(271") n = l (271")

< 2~ (1 + 2- k + roo Ckdt)(271") J2

k! ( k +1) k! ( l-k)2--k 1 + (k 1)2k ~ 2--k 1 +2 ,(271") - (271")

If k is even

k! 00

= ±2--k L n-k cos (271"nx) ,(271") n=l

k! 00 _

~ 2--k L n-k = IBk (0)1(271") n=l

5.** It is certainly necessary for instance that I should be continuous, andsufficient that I, I', f" should all be continuous since then len I~ K /n2 forsome K. More precisely, it is known that I being Lipschitz continuous oforder 0: > 1/2 (II (x) - I (y)1 ~ Mix - xl Q

for all x, y) is sufficient, butthat Lipschitz continuity of order 1/2 is not - see for instance [19, page240]. But conditions which are both necessary and sufficient are unknown.

6. With 9 (x) = (1 -lxi/a) on [-a ,a], = 0 elsewhere,

Fg (y) = 2100

9 (x) cos (271"xy) dx =21a

(1 - x/a) cos (271"xy) dx

[1 ]a 1 i a

= - (1 - x/a) sin (271"xy) + - sin (271"xy) dx7I"y 0 7I"ay 0

1 - cos (271"ay) sin2 (7I"ay)= = ---=-..,,;;..:...271"2ay2 a7l"2 y2 .

Then

is immediate from Parseval 's formula. If F I is integrable, the left side tendsto J~oo:f"J (y) dy as a -+ 00 while the right side tends to I (0), as in theproof of Theorem 7.22, and the result for x i- 0 follows in the same way,


7. The function is integrable and (Lipschitz) continuous on lR and

:Ff (y) = 2a 3 (sin (27l'ya) - 27l'ya cos (27l'ya))(2'7l'ya)

after two integrations by parts.

8.* (i) (a) Notice that f~oo 'l/J = 0 and that 'l/J is a test function by considering how fast ftoo 'l/J goes to zero as t --+ 00. (b) Given T and U

as specified, then for any ¢J, U'¢J = -U (¢J') = - [kf~oo ¢J' - TX]where X is formed from ¢J' in the same way as 'l/J is from ¢J, i.e.X(t) = f~oo (¢J' (u) + d¢Jo (u)) du and d = - f~oo ¢J'I f~oo ¢Jo = 0 so

that in fact X = ¢J. Then U'¢J = - [kf~oo ¢J' - TX] = TX = T¢J asrequired.

(ii) If Ui = U~ = T then for any ¢J, (UI - U2)¢J' = - (Ui - U~) ¢J = O.Put 'l/J (t) = c: (¢J + c¢Jo) as in (i), so that 'l/J' = ¢J + c¢Jo. It followsthat for any ¢J, (UI - U2)¢J = (UI - U2) 'l/J' - (UI - U2) (c¢Jo) = 0 -

{ (UI - U2)¢Jol f~oo ¢Jo } f~oo ¢J is a constant multiple of f~oo ¢J = Eo¢Jas required .

9. Using Definition 7.43 and the result of Example 7.41 we have for instance

E-I¢J = p -I¢J+RP-I¢J = p-I¢J + p-I R¢J

t' ¢J(t) + ¢J(-t) - 2¢J(0) dt + ('Xl ¢J(t) + ¢J (-t) dt,io t il t

O-I¢J = p-I¢J _ RP-I¢J= p-I¢J - p-I R¢J

= roo¢J(t)-¢J(-t)dtio t

and the other results are similar.

10. From Definition 7.40(iii) and Example 7.41(iii) we have

= -1 DP-2 (¢J) = ~p-2 (¢J')2 2

= ~ ( rl¢J' (x) - ¢J' (0) - X¢J" (0)dx

2 io x2

+ /00 ¢J~~x) dx _ ¢J' (0) - ¢J" (0))

11 ¢ (x) - ¢ (0) - x¢' (0) - X2¢" (0) /2

d100 ¢ (x) d

= 3 x+ 3 xo X I X

- ~ (¢J(0) +2¢J' (0) + ~¢JII (0))

B. Hints and Solutions for Exercises_________________________ 279

after another integration by parts. The general formula is of th e form

p-k (¢) = r ¢ (x) - Pk (x) dx + /00 ¢ (x) dx +~ c . (k) ¢U) (0)Jo xk xk L.J Jo 1 j=l

where Pk is the kt h Taylor polynomial of ¢ about 0, and the coefficientsCj (k) are determined from the recurrence p-k-1 = _DP-k jk.

11. If a > 0, Dpa (¢) = <P" (¢') = - Jooo xa¢' (x) dx = - [xa¢ (x)];;C +

a Jooo xa- 1¢(x) dx = apa-1 (¢) .

Hence if a < -1, a + n > 0, a +m > 0 with n > m then

1 trt-r:(a+ 1)···(a+n)

=

a + n Dn-1pa+n-1(a+l) .. · (a + n )(a+n) · .. (a+m+l) Dmpa+m

(a+l) .. · (a + n)

1 tr-t-:»(a+l) .. · (a + m ) .

Once the consistency is proved, the relation D'P" = apa-1 for other valuesof a follows from Definition 7.40(ii) or (iii) as appropriate.

12. We have

~ (Ff (y +h) - Ff (y)) = 1: f (x) e_27riXye-27ri~h -1 dx

and since for real t, lei t -11 = 2Isin(tj2)1::; ItI the integrand is dominatedby Ixf (x)1 which is given to be integrable, we can let h --+ 0 to get

DFf (y) = -21ri1: f (x) e-27rixyx dx = -21riFg (y).

13. With e > 0,

p-1-e (¢) = roo ¢ (x) - ¢ (0) dx as in Example 7.41(i)Jo xe+1

= t ¢ (x) - ¢ (0)dx + /00 ¢ (x) dx _ ~¢ (0)Jo xe+1 1 xe+1 €

so lime-+o+ p-1-e - 8j e --+ p-1 also.

14. With the given conditions, Example 7.41 (iii) gives P-2(¢) = Jo1(¢(x ) ¢(0))x-2dx - ¢(O) which is clearly < O.

15. (i) The results for a 'I 0 follow at once from Exercise 11. Also DEo (¢) =_Eo (¢') = - J~oo ¢' (x) dx = 0 and DOo is similar .


(ii) Prom (i), D (pEa) = D (oa+!) , i.e. E" + paoa-l = (a + 1) E" ; orpoa-l = E" for a4 0 and pEa-l = O" similarly. When a = 0 we havefor instance

pE- 1 (¢) = E - 1 (p¢)

(I p(x)¢(x) +p(-x) ¢(-x) - 2p(0) ¢(0) dxJo x+100 p( x)¢(x)+p(-x) ¢(-x)dx

1 x

= roo x (¢ (x) - ¢ (- x)) dx = 00 (¢) .Jo x

(iii) If pT = 0 then from Theorem 7.50(iii) it follows that DFT =

-2-rriF(pT) = 0 and so FT = kED . Hence T = kFEo = k8.

(iv) £1 = Aln-RAln.

(v) If e > 0 then

0-1+£(¢) = roo ¢ (x) - ¢ (-x) dx ~ roo ¢ (x) - ¢ (-x) dx = 0- 1 (¢)Jo x1- £ Jo x

and the other parts are similar.

16. Here eCOS X has equal maxima at 0, 2-rr and since n is an integer, the same istrue for cosnx. This contributes a factor of 2 in the result since we have twoequal maxima. Expanding eCOS X = e1-

x 2/

2+... = e (1- x 2 /2+ ...) showsthat in the formula (7.29) we have l = e, a = e/2, p = 2, b = 1, and so(after checking that the other necessary properties of f ,9 are satisfied) wehave

12~ 1 ()1/2 ~etcosx cosnx dx '" 2._ .et --.:..../ r (1/2) = et ~.° 2 te 2 t

17.* (i) an+!-bn+! = (an + bn)/2-~ < (an + bn) /2-bn = (an - bn) /2so the difference is more than halved at each step, while (bn ) increasesand (an) decreases. Hence if we take N so that aN - bN < 8bo thend = bN has the required property. Then for n ~ N,


(ii) Given

put tant = x to get

281

Write a, b for ao,bo and make the substitution bx2 = ay2 to show that

18. If X n '" end then X n +1 = In (1 + xn ) = X n -x~J2+ ... and so c = 2, d = -1by comparing terms. To make this precise, follow the method of Example7.66 using

x - ~2 < In (1 + x) < x _ (~ _ £) x 2,

11 + x < 1 _ x < 1 + (1 + e) x

for small enough x.

19. The question outlines all the necessary steps. The only point which mightbe overlooked is that the equations

0= 2C2- 282

,

1r0= - -4C8

2


determine C, S only up to sign, so it is necessary to say why the positivevalues should be chosen. But this is easy for S since

100 100 sin t 00 l(n+l)71" sin t

S = sin (x2) dx = edt = L edt

o 0 2yt 0 n71" 2yt

171" 00 (-ItsintL~dt

o 0 t+mr

which is positive since the terms of the series decrease strictly.

Bibliography

[1] T .M. Apostol. Mathematical Analysis. Addison-Wesley, Reading, MA,USA, 1957.

[2] RP. Boas (Jr.) . A Primer of Real Functions. Mat hematical Associationof America, Washington, DC, USA, 1981.

[3] H.-D. Ebbinghaus, H. Hermes, F. Hirzebruch, M. Koecher, K. Mainzer,J. Neukirch, A. Prestel, and R Remmert . Numbers. Springer, New York,NY, USA, 2nd edition, 1990.

[4] G.H. Hardy. Divergent Series. Oxford University Pr ess, Oxford , UK, 1949.

[5] E. Hewit t and K. Stromberg. Real and Abs tract Analysis. Springer , Berlin ,Germany, 1965.

[6] M.J . Lighthill. Fouri er Analysis and Generalised Functions. CambridgeUniversity Press , Cambridge, UK, 1962.

[7] G. P6lya. Mathematics and Plausible Reason ing. Princeton UniversityPr ess, Princeton, NJ, USA, 1954.

[8] G. P6lya and G. Szego. Problems and Th eorems in Analysis . Springer,New York, NY, USA, 1978.

[9] J. I. Richards and H.K. Youn. Theory of Distributions. Cambridge University Press, Cambridge, UK, 1990.

[10] W. Rudin . Principles of Mathematical Analysis. McGraw-Hill, New York,NY, USA, 2nd edition, 1964.

[11] W. Rudin. Real and Complex Analysis. McGraw-Hill, New York, NY,USA, 1966.

283


[12] W. Rudin . Functional Analysis. McGraw-Hill, New York, NY, USA, 1973.

[13] L. Schwartz. Theorie des Distributions. Hermann , Paris , France , 1957.

[14] G. Smith. Introductory Mathematics: Algebra and Analysis. Springer Un-dergraduate Mathematics Series. Springer, London, UK, 1998.

[15] P.L. Walker. A bijection from Z to Q. Math Gazette, page 119, 1995.

[16] P.L. Walker. Elliptic Functions , a Constructive Approach. J. Wiley andSons, Chichester, UK, 1996.

[17] E.T. Whittaker and G.N. Watson. A Course of Modern Analysis. Cambridge University Press , Cambridge, UK, 4th edition, 1927.

[18] R. Wong. Asymptotic Approximation of Integrals. Academic Press, London, UK, 1989.

[19] A. Zygmund. Trigonometric Serie s. Cambridge University Press, Cambridge, UK, 2nd edit ion, 1968.

Alternating Series Test, 149argument, 171arithmetic-geometric mean , 29, 245

Bernoulli funct ions, 243Bernoulli numbers, 191Bernoulli polynomials, 190, 243Bessel's inequality, 202binomial coefficient, 120bound- greatest lower, 9- least upper, 8

chain rule, 77Collatz Problem, 30comparison test, 144continuity- Lipschitz , 65cont inuous , 38- at a point , 40- on an interval, 43- on the left, 58- on the right , 58contraction, 54convergence, 9- absolute, 151- conditional, 151- general principle of, 25- of infinite products, 185- pointwise, 173- uniform, 174countable, 60critical point, 94

derivative, 68

Index

derivatives- right, left hand, 68Dirac delta-function, 219Dirichlet kernel , 205Dirichlet 's test , 150dissection, 104distribution, 221- derivative of, 222- Fourier transform of, 228- primitive of, 244dominated convergence- for integrals, 180- for products, 187- for series , 183

Euler numbers, 191Euler 's constant , 149Euler-Maclaurin formula, 236, 243Eulerian integrals, 133

first derivative test, 95Fourier coefficients, 196Fourier integrals, 208Fourier series, 195Fourier transform, 208function- bounded, 37- co-domain, 31- composition, 42- continuous, 38, 40- decreasing, 35- differentiable, 68- domain , 31- even, 122- exponential, 163

285

286

- generalised , 221- increasing, 35- integrable, 130- inverse, 52- limit of, 38- monotone, 35- nowhere differentiable, 177- odd , 122- periodic, 34- polynomial, 32- range, 31- rational, 32- regulated , 57- slowly increasing, 227- step, 104- test, 219functions- hyperbolic, 192- trigonometric, 167fundamental theorem of calculus, 112

Gamma function, 132, 133goats, 49

harmonic numbers, 148Heaviside's function , 223

improper integrals- at infinity, 128- on an interval , 126infinite products, 184inflexion- point of, 97inherited properties, 20integrable- improperly, 126, 128integrable functions, 130integral- indefinite, 114- of a regulated function, 110integral test, 147integrand, 110integration- by parts, 114- by substitution, 115inverse, 52iterated sine, 241iteration, 54

jump, 57

L' Hopital's rule, 83lemma- Abel 's , 160

Examples and Theorems in Analysis

- Rising Sun , 65limit , 10- at infinity, 59- double, 26- inferior (lim inf), 30- infinite , 59- iterated (repeated) , 26- on the left, 57- on the right , 57- one-sided, 56- recursive , 53- superior (lim sup), 30Lipschitz continuity, 203logarithm- complex , 171- real, 165

M-test, 176maximum, minimum points, 94method of bisection, 49

Newton's method, 87

orthogonality, 197

Parseval's equation, 217partial derivatives, 251partial sums, 142peak point, 21polar form, 171power series , 157powers- as distributions, 224- complex, 171- real , 166proof- by bisection, 46- by contradiction, 47

radian measure, 167radius of convergence , 160ratio test , 145refinement , 104reflection operator, 210, 226regulated function , 57

second derivative test, 97sequence , 1- bounded, 7- Cauchy, 23- complex, 158- convergent, 10- decreasing, 4- divergent, 10

Index

- double, 26- increasing, 4- monotone, 4- tends to infinity, 18series, 142- complex, 158- convergent, 142- divergent, 142- double , 155- double sum, 156- geometric, 142- harmonic, 143- iterated sum, 156- positive terms, 144- telescoping, 143stationary point, 94subsequence, 20summability- for integrals, 139- for series, 194

Taylor polynomial, 80theorem- Binomial , 120- Bolzano-Weierstrass, 63- Chebychev's , 65- de Moivre's , 172- Dini 's, 175- Fubini's, 253- Horizontal Chord, 65- Intermediate Value, 48- Mean Value, 73- Rolle's, 72- Sandwich , 15- Taylor 's, 81, 119trigonometric polynomial, 198

uncountable, 60uniform continuity, 250uniform convergence, 109

vanishing condition, 143

287

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